package 力扣;

/**
 * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 * <p>
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 * <p>
 * 此外，你可以假设该网格的四条边均被水包围。
 * <p>
 *  
 * <p>
 * 示例 1：
 * <p>
 * 输入：grid = [
 * ["1","1","1","1","0"],
 * ["1","1","0","1","0"],
 * ["1","1","0","0","0"],
 * ["0","0","0","0","0"]
 * ]
 * 输出：1
 * <p>
 * 示例 2：
 * 输入：grid = [
 * ["1","1","0","0","0"],
 * ["1","1","0","0","0"],
 * ["0","0","1","0","0"],
 * ["0","0","0","1","1"]
 * ]
 * 输出：3
 */
public class _200岛屿数量 {

    /* dfs */
    private int m, n;
    private int[][] direction = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        m = grid.length;
        n = grid[0].length;
        int ret = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j);
                    ret++;
                }
            }
        }
        return ret;
    }

    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') return;
        grid[i][j] = '0';
        //深度优先
        /* */


        for (int[] d : direction) {
            dfs(grid, i + d[0], j + d[1]);
        }
	    /* 等价于上面的那种写法
	     dfs(grid,i+1,j);
        dfs(grid,i-1,j);
        dfs(grid,i,j+1);
        dfs(grid,i,j-1);
	     */
    }
}
